Sophie Germain Prime cannot be 6n+1

Theorem

Let $p$ be a Sophie Germain prime.

Then $p$ cannot be of the form $6 n + 1$, where $n$ is a positive integer.

Proof

Let $p$ be a Sophie Germain prime.

Then, by definition, $2 p + 1$ is prime.

Aiming for a contradiction, suppose $p = 6 n + 1$ for some $n \in \Z_{>0}$.

Then:

\(\ds 2 p + 1\)

\(=\)

\(\ds 2 \paren {6 n + 1} + 1\)

\(\ds \)

\(=\)

\(\ds 12 n + 3\)

\(\ds \)

\(=\)

\(\ds 3 \paren {4 n + 1}\)

and so $2 p + 1$ is not prime.

The result follows by Proof by Contradiction.

$\blacksquare$